Bài 4: 

b: \(=x^2z\left(-1+3-7\right)=-5x^2z=-5\cdot\left(-1\right)^2\cdot\left(-2\right)=10\)

c: \(=xy^2\left(5+0.5-3\right)=2.5xy^2=2.5\cdot2\cdot1^2=5\)

Lời gải:
$A=2x+7-3=2x+4$

$B=3x-8-12=3x-20$

\(a,B=4x^2+20x+25-9+x^2+14=5x^2+20x+30\\ b,B=5\left(x^2+4x+4\right)+10\\ B=5\left(x+2\right)^2+10\ge10>0,\forall x\)

Do đó B luôn dương với mọi x

giúp

\(\left(\dfrac{1}{2}xy\right)^3.\left(\dfrac{2}{3}yt\right)^2.\left(-xy\right)\\ =\dfrac{1}{8}x^3y^3.\dfrac{4}{9}y^2t^2.\left(-1\right)xy\\ =\left[\dfrac{1}{8}.\dfrac{4}{9}.\left(-1\right)\right]\left(x^3.x\right)\left(y^3.y^2.y\right).t^2\\ =-\dfrac{1}{18}x^4y^6t^2\)

\(=\left|\sqrt{11}-3\right|-\left|\sqrt{11}-4\right|=\sqrt{11}-3+\sqrt{11}-4=-7+2\sqrt{11}\)

\(=\sqrt{11}-3-4+\sqrt{11}=-7\)

\(\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}=\sqrt{3-2\sqrt{5}+3}=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)

\(=\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}=\sqrt{3-2\sqrt{5}+3}=\sqrt{9-2\sqrt{5}}\)

\(\dfrac{2}{\sqrt{3}-1}-\dfrac{4}{1+\sqrt{5}}-\left(\sqrt{3}-\sqrt{5}\right)=\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-\dfrac{4\left(1-\sqrt{5}\right)}{1-5}-\sqrt{3}+\sqrt{5}=\sqrt{3}+1+1-\sqrt{5}-\sqrt{3}+\sqrt{5}=2\)

\(1,\\ a,=x^2-6x+8+3x^2-15x=4x^2-21x+8\\ b,=9x^2+12x+4-x^2+9=8x^2+12x+13\\ 2,\\ a,\Leftrightarrow x^2+8x+16-x^2+4=5\\ \Leftrightarrow8x=-15\Leftrightarrow x=-\dfrac{15}{8}\\ b,\Leftrightarrow9x^2-6x+1-8x^2-2x+12x+3-x^2=5\\ \Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\)